Units & Dimensions

JEE Main level, single correct, medium

4 of 149

Question 4 of 149

Match List-I with List-II.
List-IList-II(A) Heat capacity of body(I) Jkg−1(B) Specific heat capacity of body(II) JK−1(C) Latent heat(III) Jkg−1 K−1(D) Thermal conductivity(IV) Jm−1 K−1 s−1\begin{array}{ll} \text{List-I} & \text{List-II} \\ \text{(A) Heat capacity of body} & \text{(I) } \mathrm{J} \mathrm{kg}^{-1} \\ \text{(B) Specific heat capacity of body} & \text{(II) } \mathrm{JK}^{-1} \\ \text{(C) Latent heat} & \text{(III) } \mathrm{J} \mathrm{kg}^{-1} \mathrm{~K}^{-1} \\ \text{(D) Thermal conductivity} & \text{(IV) } \mathrm{Jm}^{-1} \mathrm{~K}^{-1} \mathrm{~s}^{-1} \end{array}
Choose the correct answer from the options given below :

Show answer and solution

Answer: (D) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

C′=ΔQΔT=JK−1 S=ΔQmΔT=Jkg−1 K−1 L=ΔQm=Jkg−1ΔQ=KAΔTL⇒K=ΔQ(L)AΔT=Jm−1 K−1 s−1\begin{aligned} & \mathrm{C}^{\prime}=\frac{\Delta \mathrm{Q}}{\Delta \mathrm{T}}=\mathrm{JK}^{-1} \\ & \mathrm{~S}=\frac{\Delta \mathrm{Q}}{\mathrm{m} \Delta \mathrm{T}}=\mathrm{Jkg}^{-1} \mathrm{~K}^{-1} \\ & \mathrm{~L}=\frac{\Delta \mathrm{Q}}{\mathrm{m}}=\mathrm{Jkg}^{-1} \\ & \Delta \mathrm{Q}=\frac{\mathrm{KA} \Delta \mathrm{T}}{\mathrm{L}} \Rightarrow \mathrm{K}=\frac{\Delta \mathrm{Q}(\mathrm{L})}{\mathrm{A} \Delta \mathrm{T}}=\mathrm{Jm}^{-1} \mathrm{~K}^{-1} \mathrm{~s}^{-1}\end{aligned}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Units & Dimensions (Units), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.